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UA OPTI501 电磁波 求解麦克斯韦方程组的Fourier方法3 Coulomb Gauge下讨论Maxwell方程

发布时间：2025/4/14 编程问答 55 豆豆
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UA OPTI501 电磁波 求解麦克斯韦方程组的Fourier方法3 Coulomb Gauge下讨论Maxwell方程

Use the macroscopic equation with D-field, H-field eliminated by bound charge and bound current:
∇⋅E=ρtotal(e)ϵ0∇×B=μ0Jtotal(e)+μ0ϵ0∂∂tE∇×E=−∂∂tB∇⋅B=0\nabla \cdot \textbf E =\frac{\rho_{total}^{(e)}}{\epsilon_0} \\ \nabla \times \textbf B=\mu_0 \textbf J^{(e)}_{total}+\mu_0 \epsilon_0\frac{\partial}{\partial t}\textbf E \\ \nabla \times \textbf E=-\frac{\partial}{\partial t}\textbf B \\ \nabla \cdot \textbf B=0∇⋅E=ϵ0​ρtotal(e)​​∇×B=μ0​Jtotal(e)​+μ0​ϵ0​∂t∂​E∇×E=−∂t∂​B∇⋅B=0

Also define scalar potential ψ\psiψ and A\textbf AA satisfying
B=∇×AE=−∇ψ−∂A∂t\textbf B = \nabla \times \textbf A \\ \textbf E = -\nabla \psi - \frac{\partial \textbf A}{\partial t}B=∇×AE=−∇ψ−∂t∂A​

so that Maxwell 3 and 4 are naturally satisfied. Now assume
ρtotal(e)=ρ0ei(k⋅r−wt),Jtotal(e)=J0ei(k⋅r−wt)A=(A∣∣+A⊥)ei(k⋅r−wt),ψ=ψ0ei(k⋅r−wt)\rho^{(e)}_{total}=\rho_0e^{i(\textbf k\cdot \textbf r - wt)}, \textbf J_{total}^{(e)}=\textbf J_0 e^{i(\textbf k \cdot \textbf r -wt)} \\ \textbf A = (\textbf A_{||}+\textbf A_{\perp})e^{i(\textbf k \cdot \textbf r - wt)},\psi = \psi_0 e^{i(\textbf k \cdot \textbf r - wt)}ρtotal(e)​=ρ0​ei(k⋅r−wt),Jtotal(e)​=J0​ei(k⋅r−wt)A=(A∣∣​+A⊥​)ei(k⋅r−wt),ψ=ψ0​ei(k⋅r−wt)

Under Coulomb gauge, ∇⋅A=0\nabla \cdot \textbf A=0∇⋅A=0. Replace ∇\nabla∇ by iki \textbf kik and ∂∂t\frac{\partial}{\partial t}∂t∂​ by −iw-iw−iw, ik⋅(A∣∣+A⊥)ei(k⋅r−wt)=0i\textbf k \cdot (\textbf A_{||}+\textbf A_{\perp})e^{i(\textbf k \cdot \textbf r - wt)}=0ik⋅(A∣∣​+A⊥​)ei(k⋅r−wt)=0, A∣∣=0\textbf A_{||}=0A∣∣​=0. So A=A⊥ei(k⋅r−wt)\textbf A = \textbf A_{\perp}e^{i(\textbf k \cdot \textbf r - wt)}A=A⊥​ei(k⋅r−wt). As a result,
B=(B∣∣+B⊥)ei(k⋅r−wt)=ik×A⊥ei(k⋅r−wt)B∣∣=0,B⊥=ik×A⊥E=(E∣∣+E⊥)ei(k⋅r−wt)=−ikψ0ei(k⋅r−wt)+iwA⊥ei(k⋅r−wt)E∣∣=−ikψ0,E⊥=iwA⊥\textbf B =( \textbf B_{||}+\textbf B_{\perp}) e^{i(\textbf k \cdot \textbf r - wt)}= i \textbf k \times \textbf A_{\perp}e^{i(\textbf k \cdot \textbf r - wt)} \\ \textbf B_{||}=0,\textbf B_{\perp} = i \textbf k \times \textbf A_{\perp} \\ \textbf E =( \textbf E_{||}+\textbf E_{\perp}) e^{i(\textbf k \cdot \textbf r - wt)}=-i \textbf k \psi_0 e^{i(\textbf k \cdot \textbf r - wt)} +iw\textbf A_{\perp}e^{i(\textbf k \cdot \textbf r - wt)} \\ \textbf E_{||} = -i \textbf k \psi_0, \textbf E_{\perp} =iw\textbf A_{\perp} B=(B∣∣​+B⊥​)ei(k⋅r−wt)=ik×A⊥​ei(k⋅r−wt)B∣∣​=0,B⊥​=ik×A⊥​E=(E∣∣​+E⊥​)ei(k⋅r−wt)=−ikψ0​ei(k⋅r−wt)+iwA⊥​ei(k⋅r−wt)E∣∣​=−ikψ0​,E⊥​=iwA⊥​

Now replace ∇\nabla∇ by iki \textbf kik and ∂∂t\frac{\partial}{\partial t}∂t∂​ by −iw-iw−iw in Maxwell 1 and 2.
ik⋅(E∣∣+E⊥)=k2ψ0=ρ0ϵ0,ψ0=ρ0ϵ0k2ik×B⊥=μ0J0−iwμ0ϵ0(E∣∣+E⊥)i \textbf k \cdot( \textbf E_{||}+\textbf E_{\perp}) =k^2 \psi_0=\frac{\rho_0}{\epsilon_0},\psi_0 = \frac{\rho_0}{\epsilon_0 k^2} \\ i \textbf k \times \textbf B_{\perp} = \mu_0\textbf J_0 -iw\mu_0 \epsilon_0 ( \textbf E_{||}+\textbf E_{\perp}) ik⋅(E∣∣​+E⊥​)=k2ψ0​=ϵ0​ρ0​​,ψ0​=ϵ0​k2ρ0​​ik×B⊥​=μ0​J0​−iwμ0​ϵ0​(E∣∣​+E⊥​)

LHS is ik×(ik×A⊥)=−k×(k×A⊥)=k2A⊥\begin{aligned} i \textbf k \times(i \textbf k \times \textbf A_{\perp}) & = - \textbf k \times(\textbf k \times \textbf A_{\perp}) = k^2 \textbf A_{\perp}\end{aligned}ik×(ik×A⊥​)​=−k×(k×A⊥​)=k2A⊥​​ and the second term of RHS is iwμ0ϵ0(E∣∣+E⊥)=iwμ0ϵ0(−ikψ0+iwA⊥)=μ0wρ0k/k2−(w/c)2A⊥iw\mu_0 \epsilon_0 ( \textbf E_{||}+\textbf E_{\perp}) =iw\mu_0 \epsilon_0 ( -i \textbf k \psi_0+iw\textbf A_{\perp} )=\mu_0w\rho_0 \textbf k/k^2 -(w/c)^2\textbf A_{\perp}iwμ0​ϵ0​(E∣∣​+E⊥​)=iwμ0​ϵ0​(−ikψ0​+iwA⊥​)=μ0​wρ0​k/k2−(w/c)2A⊥​. Thus,
k2A⊥=μ0(J0∣∣+J0⊥)−μ0wρ0k/k2+(w/c)2A⊥k^2 \textbf A_{\perp} = \mu_0(\textbf J_{0 ||}+\textbf J_{0 \perp})-\mu_0w\rho_0 \textbf k/k^2+(w/c)^2\textbf A_{\perp}k2A⊥​=μ0​(J0∣∣​+J0⊥​)−μ0​wρ0​k/k2+(w/c)2A⊥​

Consider the continuity equation of charge,
ik⋅J0−iwρ0=0J0∣∣−wρk^/k=0i \textbf k \cdot \textbf J_0 -i w \rho_0 = 0 \\ \textbf J_{0||}-w\rho \hat k/k=0ik⋅J0​−iwρ0​=0J0∣∣​−wρk^/k=0

Thus,
k2A⊥=μ0J0⊥+(w/c)2A⊥A⊥=μ0J⊥k2−(w/c)2k^2 \textbf A_{\perp} = \mu_0\textbf J_{0 \perp}+(w/c)^2\textbf A_{\perp} \\ \textbf A_{\perp} = \frac{\mu_0 \textbf J_{\perp}}{k^2-(w/c)^2}k2A⊥​=μ0​J0⊥​+(w/c)2A⊥​A⊥​=k2−(w/c)2μ0​J⊥​​

So the E-field and B-field are
E=−ikψ0+iwA⊥=−iρ0kϵ0k2+iμ0wJ0⊥k2−(w/c)2B=ik×A⊥=iμ0k×J0k2−(w/c)2\textbf E = -i \textbf k \psi_0+iw\textbf A_{\perp}=-i \frac{\rho_0 \textbf k}{\epsilon_0 k^2}+i\frac{\mu_0 w \textbf J_{0 \perp}}{k^2-(w/c)^2} \\ \textbf B = i \textbf k \times \textbf A_{\perp}=i\frac{\mu_0 \textbf k \times \textbf J_0}{k^2-(w/c)^2}E=−ikψ0​+iwA⊥​=−iϵ0​k2ρ0​k​+ik2−(w/c)2μ0​wJ0⊥​​B=ik×A⊥​=ik2−(w/c)2μ0​k×J0​​

B-field is the same as the result under Lorenz gauge. However, the E-field looks a little different. The E-field under Lorenz gauge is
E=−i(ρ0/ϵ0)k+iμ0wJ0k2−(w/c)2\textbf E=\frac{-i(\rho_0/\epsilon_0)\textbf k+i\mu_0 w \textbf J_0}{k^2-(w/c)^2}E=k2−(w/c)2−i(ρ0​/ϵ0​)k+iμ0​wJ0​​

Since
J0=J0∣∣+J0⊥=wρk/k2+J0⊥\textbf J_0=\textbf J_{0 ||}+\textbf J_{0 \perp}=w\rho \textbf k/k^2+\textbf J_{0 \perp}J0​=J0∣∣​+J0⊥​=wρk/k2+J0⊥​

The E-field under Lorenz field can be reformulated as
E=−i(ρ0/ϵ0)k+iμ0w2ρk/k2+iμ0wJ0⊥k2−(w/c)2=−iρ0kϵ0k2+iμ0wJ0⊥k2−(w/c)2\textbf E=\frac{-i(\rho_0/\epsilon_0)\textbf k+i\mu_0w^2\rho \textbf k/k^2+i\mu_0 w \textbf J_{0 \perp}}{k^2-(w/c)^2}=-i \frac{\rho_0 \textbf k}{\epsilon_0 k^2}+i\frac{\mu_0 w \textbf J_{0 \perp}}{k^2-(w/c)^2}E=k2−(w/c)2−i(ρ0​/ϵ0​)k+iμ0​w2ρk/k2+iμ0​wJ0⊥​​=−iϵ0​k2ρ0​k​+ik2−(w/c)2μ0​wJ0⊥​​

which is the same as the E-field under Couloub gauge.

总结

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